Editing Acoustic theory (section)

===For a moving medium===
Again, we can derive the small-disturbance limit for sound waves in a moving medium. Again, starting with
: <math>
  \begin{align}
     \frac{\partial \rho'}{\partial t} +\rho_0\nabla\cdot\mathbf{v}+\mathbf{u}\cdot\nabla\rho' + \nabla\cdot \rho'\mathbf{v} & = 0 \\
     (\rho_0+\rho')\frac{\partial \mathbf{v}}{\partial t} + (\rho_0+\rho')(\mathbf{u}\cdot\nabla)\mathbf{v} + (\rho_0+\rho')(\mathbf{v}\cdot\nabla)\mathbf{v} + \nabla p' & = 0
  \end{align}
</math>

We can linearize these into
: <math>
  \begin{align}
     \frac{\partial \rho'}{\partial t} +\rho_0\nabla\cdot\mathbf{v}+\mathbf{u}\cdot\nabla\rho' & = 0 \\
     \frac{\partial \mathbf{v}}{\partial t} + (\mathbf{u}\cdot\nabla)\mathbf{v} + \frac{1}{\rho_0}\nabla p' & = 0
  \end{align}
</math>

====For Irrotational Fluids in a Moving Medium====
Given that we saw that
: <math>
  \begin{align}
     \frac{\partial \rho'}{\partial t} +\rho_0\nabla\cdot\mathbf{v}+\mathbf{u}\cdot\nabla\rho' & = 0 \\
     \frac{\partial \mathbf{v}}{\partial t} + (\mathbf{u}\cdot\nabla)\mathbf{v} + \frac{1}{\rho_0}\nabla p' & = 0
  \end{align}
</math>

If we make the previous assumptions of the fluid being ideal and the velocity being irrotational, then we have
: <math>
  \begin{align}
     p' & = \left(\frac{\partial p}{\partial \rho_{0}}\right)_{s}\rho' = c^{2}\rho' \\
     \mathbf{v} & = -\nabla\phi
  \end{align}
</math>

Under these assumptions, our linearized sound equations become
: <math>
  \begin{align}
     \frac{1}{c^2}\frac{\partial p'}{\partial t} -\rho_0\nabla^2\phi+\frac{1}{c^2}\mathbf{u}\cdot\nabla p' & = 0 \\
     -\frac{\partial}{\partial t}(\nabla\phi) - (\mathbf{u}\cdot\nabla)[\nabla\phi] + \frac{1}{\rho_0}\nabla p' & = 0
  \end{align}
</math>

Importantly, since <math>\mathbf{u}</math> is a constant, we have <math>(\mathbf{u}\cdot\nabla)[\nabla\phi] = \nabla[(\mathbf{u}\cdot\nabla)\phi]</math>, and then the second equation tells us that
: <math>
  \frac{1}{\rho_0} \nabla p' = \nabla\left[\frac{\partial\phi}{\partial t} + (\mathbf{u}\cdot\nabla)\phi\right]
</math>

Or just that
: <math>
  p' = \rho_{0}\left[\frac{\partial\phi}{\partial t} + (\mathbf{u}\cdot\nabla)\phi\right]
</math>

Now, when we use this relation with the fact that <math>\frac{1}{c^2}\frac{\partial p'}{\partial t} -\rho_0\nabla^2\phi+\frac{1}{c^2}\mathbf{u}\cdot\nabla p' = 0</math>, alongside cancelling and rearranging terms, we arrive at
: <math>
  \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} - \nabla^2\phi + \frac{1}{c^2}\frac{\partial}{\partial t}[(\mathbf{u}\cdot\nabla)\phi] + \frac{1}{c^2}\frac{\partial}{\partial t}(\mathbf{u}\cdot\nabla\phi) + \frac{1}{c^2}\mathbf{u}\cdot\nabla[(\mathbf{u}\cdot\nabla)\phi] = 0
</math>

We can write this in a familiar form as
:<math>
\left[\frac{1}{c^2}\left(\frac{\partial}{\partial t} + \mathbf{u}\cdot\nabla\right)^{2} - \nabla^{2}\right]\phi = 0
</math>

This differential equation must be solved with the appropriate boundary conditions. Note that setting <math>\mathbf{u}=0</math> returns us the wave equation. Regardless, upon solving this equation for a moving medium, we then have
:<math>
\begin{align}
   \mathbf{v} & = -\nabla \phi \\
p' & = \rho_{0}\left(\frac{\partial}{\partial t} + \mathbf{u}\cdot\nabla\right)\phi\\
\rho' & = \frac{\rho_{0}}{c^{2}}\left(\frac{\partial}{\partial t} + \mathbf{u}\cdot\nabla\right)\phi
\end{align}
 </math>